POJ1308：Is It A Tree?（并查集）

Is It A Tree?

题目链接：http://poj.org/problem?id=1308

Description:

A tree is a well-known data structure that is either empty (null, void, nothing) or is a set of one or more nodes connected by directed edges between nodes satisfying the following properties.

There is exactly one node, called the root, to which no directed edges point.
Every node except the root has exactly one edge pointing to it.
There is a unique sequence of directed edges from the root to each node.
For example, consider the illustrations below, in which nodes are represented by circles and edges are represented by lines with arrowheads. The first two of these are trees, but the last is not.

In this problem you will be given several descriptions of collections of nodes connected by directed edges. For each of these you are to determine if the collection satisfies the definition of a tree or not.

Input:
The input will consist of a sequence of descriptions (test cases) followed by a pair of negative integers. Each test case will consist of a sequence of edge descriptions followed by a pair of zeroes Each edge description will consist of a pair of integers; the first integer identifies the node from which the edge begins, and the second integer identifies the node to which the edge is directed. Node numbers will always be greater than zero.

Output:
For each test case display the line "Case k is a tree." or the line "Case k is not a tree.", where k corresponds to the test case number (they are sequentially numbered starting with 1).

Sample Input:
6 8  5 3  5 2  6 4
5 6  0 0

8 1  7 3  6 2  8 9  7 5
7 4  7 8  7 6  0 0

3 8  6 8  6 4
5 3  5 6  5 2  0 0
-1 -1

Sample Output:
Case 1 is a tree.
Case 2 is a tree.
Case 3 is not a tree.

题意：
给出多对有向边，问是否能构成一棵树。树的定义：无环，不是森林且每个点（除开根节点）入度为1。另外，如果为空也是一棵树。

题解：
把题目要求分析清楚后，发现除开入度问题其它都可以用并查集解决，并查集可以很轻松地判断是否有环以及是否为一颗森林。

统计入度很简单，开个数组记录一下就好了。

代码如下:

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <iostream>
using namespace std;

const int N = 1005;
int f[N],in[N],vis[N];
int find(int x){
    return f[x]==x ? x : f[x]=find(f[x]);
}
int main(){
    int u,v,end=0,tot=0;
    while(1){
        tot++;
        memset(in,0,sizeof(in));memset(vis,0,sizeof(vis));
        for(int i=1;i<=N-5;i++) f[i]=i;
        bool flag = true;
        while(cin>>u>>v){
            if(u==-1 && v==-1) end=1;
            if(u<=0 && v<=0) break;
            in[v]++;vis[u]=vis[v]=1;
            if(in[v]>1) flag=false;
            int fu=find(u),fv=find(v);
            if(fu!=fv) f[fu]=fv;
            else flag=false;
        }
        if(end) break;
        int cnt = 0;
        for(int i=1;i<=N-5;i++) if(f[i]==i &&vis[i]) cnt++;
        if(cnt>1) flag=false;
        if(!flag) printf("Case %d is not a tree.
",tot);
        else printf("Case %d is a tree.
",tot);
    }
    return 0;
}